Question

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is initially traveling at speed 7.0v? Assume that the acceleration due to the braking is the same in both cases.
Express your answer using two significant figures.

Answer 1

49d

Further explanation

This case is about uniformly accelerated motion.

Given:

The initial speed was v takes distance d to stop after the brakes are applied.

Question:

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

The Process:

The list of variables to be considered is as follows.

• $\boxed{u \ or \ v_i = initial \ velocity}$
• $\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
• $\boxed{a = acceleration \ (constant)}$
• $\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

• a = acceleration (in m/s²)
• u = initial velocity  
• v = final velocity
• d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

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Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

Answer 2

The stopping distance if the car is initially travelling at speed $7.0v$ is $\boxed{49d}$.

Further Explanation:

The term motion is defined as the change in the position of the object with respect to the surrounding.

There are three main equations of motion also known as laws of constant acceleration because these equations are valid only when the body has uniform acceleration.

These equations used to find the variable like initial and final velocity, acceleration and time.

Let u is the initial velocity,$v$ is the final velocity, $a$ is the acceleration, $s$ is the distance and $t$ is the time.

The first equation of motion is:

$\boxed{v=u+at}$                                                          …… (1)

The second equation of motion is:

$\boxed{v^{2}=u^2+2as}$                                            …… (2)

The third equation of motion is:

$\boxed{s=ut+\dfrac{1}{2}{at^2}}$                                 …… (3)

Given:

The speed of the car is v and travel the distance d to stop after breaks applied.

The new speed at which car is traveling is $0.7v$.

Concept:

Apply the second equation of motion.

Case1:

The initial velocity of the car is $v$, the final velocity of the car is $0$, the distance covered by the car after break is applied is $d$ and the acceleration during break is $- a$.

Substitute $v$ for $u$, $0$ for $v$ and $d$ for $s$ in equation (2).

$\begin{aligned}{\left( 0 \right)^2}&={\left(v\right)^2}+2\left({-a} \right)d\hfill\\0&={v^2}-2ad\hfill\\\end{aligned}$

Simplify further.

${v^2}=2ad$                                                                  …… (4)

Case2:

The initial velocity of the car is $0.7v$, the final velocity of the car is $0$, the distance covered by the car after break is applied is $d'$ and the acceleration during break is  $- a$.

Substitute $v$ for $u$, $0$ for $v$ and $d'$ for $s$ in equation (2).

$\begin{aligned}{\left(0\right)^2}&={\left({0.7v}\right)^2}+2\left( {-a}\right)d'\hfill\\0&=49{v^2}-2ad'\hfill\\\end{aligned}$

Simplify further.

$49{v^2}=2ad'$                                                            …… (5)

Divide equation (4) by equation (5).

$\begin{aligned}\frac{{{v^2}}}{{49.0{v^2}}}&=\frac{{2ad}}{{2ad'}}\hfill\\\frac{1}{{49.0}}&=\frac{d}{{d'}}\hfill\\d'&=49d\hfill\\\end{aligned}$

Therefore, the stopping distance if the car is initially travelling at speed $7.0v$ is $49d$.

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Answer Details:

Grade: High school

Subject: Physics

Chapter: Kinematics

Keywords:

Car, travelling, speed v distance, d, stop, breaks, applied, stopping, distance, initially, traveling, 7.0v, acceleration, breaking, same, both case.